Comments on: Rock, Paper, Scissorhttps://blog.tanyakhovanova.com/2010/08/rock-paper-scissor/
Mathematics, applications of mathematics to life in general, and my life as a mathematician.Mon, 21 Jan 2019 00:03:44 +0000hourly1https://wordpress.org/?v=5.0.3By: vamsihttps://blog.tanyakhovanova.com/2010/08/rock-paper-scissor/#comment-1575
Mon, 23 Aug 2010 10:47:32 +0000https://blog.tanyakhovanova.com/?p=266#comment-1575I just observed that the scissor in the picture shown is not exactly a scissor but a half-scissor
]]>By: Jameshttps://blog.tanyakhovanova.com/2010/08/rock-paper-scissor/#comment-1574
Sat, 21 Aug 2010 00:18:15 +0000https://blog.tanyakhovanova.com/?p=266#comment-1574Hi Bill: I think your Rock-Paper-Scissor solution (each plays paper with probability 2/3) is correct if the objective is to maximise expected gain on a single turn.

However, if as Tanya described it the game goes on until the first time someone wins a turn, and if the aim is to maximise expected gain over the whole game, then the situation is different. Now the unrestricted player has a crushing advantage (given enough patience!). Suppose that you are the unrestricted player and on each turn you choose paper with probability 0.99999999 and scissors with probability 0.0000001. My best strategy in response is actually to play rock all the time! (in which case the game will definitely end on the first turn). If I sometimes choose paper, I do even worse (although I might be able to make the game to last a looooooooooooooooong time!). Essentially, as the unrestricted player, you can get as close to a sure win as you desire.

]]>By: CalcDavehttps://blog.tanyakhovanova.com/2010/08/rock-paper-scissor/#comment-1573
Thu, 19 Aug 2010 18:33:55 +0000https://blog.tanyakhovanova.com/?p=266#comment-1573Something like “Electric Paper, Rock, Scissors” where it takes a turn to recharge the last throw?
]]>By: Billhttps://blog.tanyakhovanova.com/2010/08/rock-paper-scissor/#comment-1572
Thu, 19 Aug 2010 18:11:17 +0000https://blog.tanyakhovanova.com/?p=266#comment-1572My prediction for the second game is that it will never produce a winner among players using the optimal strategy. If the first round is a draw, the game is a draw. If one player wins the first round, the players would reason as follows:

Player 1: I played Paper and beat your Rock. You can’t play Rock on the next round, so I will be safe playing Scissors. I am not allowed to play Paper, and if I play Rock, you might play Paper and beat me, tying the score. I am now winning 1-0, so if we draw on the next round I will win. Therefore, playing Scissors is the only reasonable choice.

Player 2: I played Rock and you beat me with Paper. You can’t play Paper on the next round, and so I would be safe with Rock, but unfortunately the rules prevent me from playing it. You will certainly play Scissors on the next round, so I can’t play Scissors either, or we will draw and you will win the game. I therefore must play Paper and give up the point.

The reasoning will be similar on the next round as well. Player 1 will thus be able to rack up as many points as desired, but never cash them in. The player who is behind can always avoid a game-ending draw by choosing what the other player chose on the last round, and so nobody can ever force a win.

The player who has freedom to choose the Scissor should never choose Rock, since it will never win. So one player is choosing between Scissor and Paper and the other is choosing between Paper and Rock.

Both players should choose Paper two out of every three times – or more accurately, with a 2/3 probability, as the deviations from Paper must be completely unpredictable.

This will lead to the player choosing between Paper and Scissor winning an average of one more time than the player choosing between Paper and Rock every three games. Neither player can do any better.